Limits - Substitution
As x approaches 3 from both directions, y approaches 8
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y = 2x + 2
-4
-2
0
2
4
6
8
10
12
14
16
-2 -1 0 1 2 3 4 5 6€
limx→3
2x + 2 = 8
We can find the limit by substituting x = 3 into
the equation
Practice
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limx→−2
x 2 − 9x + 2( )
Answer: The limit is 24
When we try to substitute into we get which is undefined.
If we draw the graph we find that we get a straight line with equation
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limx→1
x 2 −1x −1
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00
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y = x +1 but x ≠1 i.e. there is a "hole" in the graph.
The hole in the graph at x = 1 is a discontinuity. y has a value for every x except x = 1. i.e.
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f (1) does not exist
You can recognise a discontinuity because you need to lift your pen to continue
your graph. The graph below is continuous because we can draw it without having to lift the
pen.
Although , we do have a limit at x = 1.
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f (1) does not exist
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The limit is 2
Two methods to find the limit.Method 1
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Factorise x2 −1x −1
= x +1( ) x −1( )x −1
= x +1 under the condition that x ≠1
Now substitute x = 1 to get a limit of 2i.e.
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limx→1
x 2 −1x −1
= limx→1
x −1( )(x +1)x −1
= limx→1
(x +1) = 2
Method 2Use L’Hospital’s Rule
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Differentiate the top and bottom of x2 −1x −1
separately i.e. = 2x1
and then substitute x =1.
i.e. limx→1
x 2 −1x −1
= limx→1
2x1
= 2
Note: Only use this when substitution gives 0/0
Practice
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limx→−4
x + 416 − x 2
Answer: Substituting gives
Using either factorising or L’Hospital’s Rule:Limit is
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00
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18
Discontinuities and limits
f(0.5) = 3 (Solid dot gives the value at 0.5)
But
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limx→0.5
f (x) = 2.82
Not all discontinuities have a
limit
Jump discontinuity
The graph is not heading towards the
same value so there is no limit.
Tends towards 1
Tends towards -1
Limit at x = -4 does not exist
Vertical asymptotoes: Limit does not exist.
Note: this is NOT a discontinuityf(1) =2 and
0
1
2
3
-2 -1 0 1 2 3 4
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limx→1f (x) = 2
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limx→∞
3x − 42x + 5
More Limits
Divide top and bottom by x
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limx→∞
3 − 4x
2 + 5x
= 32
When bottom power is greater than top power
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limx→∞
3x 2 − 42x 3 + 5
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limx→∞
3x 2 − 42x 3 + 5
= 0
When top power is greater than bottom
power
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limx→∞
3x 3 − 42x 2 + 5
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limx→∞
3x 3 − 42x 2 + 5
⇒ no limit
Topic 1 Reference Page Exercise
Delta 67, 69,72, 73
6.1, 6.2,6.3, 6.5
Sidebotham 51, 58 3.1, 3.2
NuLake 5, 41 1.1, 1.11
Theta (red) 76 7.2
• identif y features of given graphs:limi ts, differentiabilit y, discontinuity,gradients, concavity
• interpret features of graph
SeniorMathematics 159 1A, 1B
Worksheet 1
What do you think the limit is at x = 0?
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